1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
|
## Copyright (C) 2011, 2012, 2016, 2018 Moreno Marzolla
##
## This file is part of the queueing toolbox.
##
## The queueing toolbox is free software: you can redistribute it and/or
## modify it under the terms of the GNU General Public License as
## published by the Free Software Foundation, either version 3 of the
## License, or (at your option) any later version.
##
## The queueing toolbox is distributed in the hope that it will be
## useful, but WITHOUT ANY WARRANTY; without even the implied warranty
## of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
## General Public License for more details.
##
## You should have received a copy of the GNU General Public License
## along with the queueing toolbox. If not, see <http://www.gnu.org/licenses/>.
## -*- texinfo -*-
##
## @deftypefn {Function File} {@var{M} =} dtmcfpt (@var{P})
##
## @cindex first passage times
## @cindex mean recurrence times
## @cindex discrete time Markov chain
## @cindex Markov chain, discrete time
## @cindex DTMC
##
## Compute mean first passage times and mean recurrence times
## for an irreducible discrete-time Markov chain over the state space
## @math{@{1, @dots{}, N@}}.
##
## @strong{INPUTS}
##
## @table @code
##
## @item @var{P}(i,j)
## transition probability from state @math{i} to state @math{j}.
## @var{P} must be an irreducible stochastic matrix, which means that
## the sum of each row must be 1 (@math{\sum_{j=1}^N P_{i j} = 1}),
## and the rank of @var{P} must be @math{N}.
##
## @end table
##
## @strong{OUTPUTS}
##
## @table @code
##
## @item @var{M}(i,j)
## For all @math{1 @leq{} i, j @leq{} N}, @math{i \neq j}, @code{@var{M}(i,j)} is
## the average number of transitions before state @var{j} is entered
## for the first time, starting from state @var{i}.
## @code{@var{M}(i,i)} is the @emph{mean recurrence time} of state
## @math{i}, and represents the average time needed to return to state
## @var{i}.
##
## @end table
##
## @strong{REFERENCES}
##
## @itemize
## @item Grinstead, Charles M.; Snell, J. Laurie (July
## 1997). @cite{Introduction to Probability}, Ch. 11: Markov
## Chains. American Mathematical Society. ISBN 978-0821807491.
## @end itemize
##
## @seealso{ctmcfpt}
##
## @end deftypefn
## Author: Moreno Marzolla <moreno.marzolla(at)unibo.it>
## Web: http://www.moreno.marzolla.name/
function result = dtmcfpt( P )
if ( nargin != 1 )
print_usage();
endif
[N err] = dtmcchkP(P);
( N>0 ) || ...
error(err);
if ( any(diag(P) == 1) )
error("Cannot compute first passage times for absorbing chains");
endif
w = dtmc(P); # steady state probability vector
W = repmat(w,N,1);
## Z = (I - P + W)^-1 where W is the matrix where each row is the
## steady-state probability vector for P
Z = inv(eye(N)-P+W);
## m_ij = (z_jj - z_ij) / w_j
result = (repmat(diag(Z)',N,1) - Z) ./ repmat(w,N,1) + diag(1./w);
endfunction
%!test
%! P = [1 1 1; 1 1 1];
%! fail( "dtmcfpt(P)" );
%!test
%! P = dtmcbd([1 1 1], [0 0 0] );
%! fail( "dtmcfpt(P)", "absorbing" );
%!test
%! P = [ 0.0 0.9 0.1; ...
%! 0.1 0.0 0.9; ...
%! 0.9 0.1 0.0 ];
%! p = dtmc(P);
%! M = dtmcfpt(P);
%! assert( diag(M)', 1./p, 1e-8 );
## Example on p. 461 of [GrSn97]
%!test
%! P = [ 0 1 0 0 0; ...
%! .25 .0 .75 0 0; ...
%! 0 .5 0 .5 0; ...
%! 0 0 .75 0 .25; ...
%! 0 0 0 1 0 ];
%! M = dtmcfpt(P);
%! assert( M, [16 1 2.6667 6.3333 21.3333; ...
%! 15 4 1.6667 5.3333 20.3333; ...
%! 18.6667 3.6667 2.6667 3.6667 18.6667; ...
%! 20.3333 5.3333 1.6667 4 15; ...
%! 21.3333 6.3333 2.6667 1 16 ], 1e-4 );
%!test
%! sz = 10;
%! P = reshape( 1:sz^2, sz, sz );
%! normP = repmat(sum(P,2),1,columns(P));
%! P = P./normP;
%! M = dtmcfpt(P);
%! for i=1:rows(P)
%! for j=1:columns(P)
%! assert( M(i,j), 1 + dot(P(i,:), M(:,j)) - P(i,j)*M(j,j), 1e-8);
%! endfor
%! endfor
## "Rat maze" problem (p. 453 of [GrSn97]);
%!test
%! P = zeros(9,9);
%! P(1,[2 4]) = .5;
%! P(2,[1 5 3]) = 1/3;
%! P(3,[2 6]) = .5;
%! P(4,[1 5 7]) = 1/3;
%! P(5,[2 4 6 8]) = 1/4;
%! P(6,[3 5 9]) = 1/3;
%! P(7,[4 8]) = .5;
%! P(8,[7 5 9]) = 1/3;
%! P(9,[6 8]) = .5;
%! M = dtmcfpt(P);
%! assert( M(1:9 != 5,5)', [6 5 6 5 5 6 5 6], 100*eps );
|
